MHTCET 25 Apr 2025 Shift 1 Paper

Concept: Use implicit differentiation twice on the equation of an ellipse to find $\frac{d^2 y}{dx^2}$ in terms of $x$ and $y$.Explanation:Start with the given equation: $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$Multiply throughout by $a^{2}b^{2}$ to simplify: $$b^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2}$$Differentiate both sides with respect to $x$: $$2b^{2}x + 2a^{2}y\frac{dy}{dx} = 0$$Solve for the first derivative: $$\frac{dy}{dx} = -\frac{b^{2}x}{a^{2}y}$$Differentiate again with respect to $x$ using the quotient rule: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{2}}{a^{2}}\left(\frac{y - x\frac{dy}{dx}}{y^{2}}\right)$$Substitute the expression for $\frac{dy}{dx}$: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{2}}{a^{2}y^{2}}\left(y - x\left(-\frac{b^{2}x}{a^{2}y}\right)\right)$$Simplify the bracket: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{2}}{a^{2}y^{2}}\left(y + \frac{b^{2}x^{2}}{a^{2}y}\right)$$Combine terms over a common denominator: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{2}}{a^{2}y^{2}} \cdot \frac{a^{2}y^{2} + b^{2}x^{2}}{a^{2}y}$$From the simplified original equation, $a^{2}y^{2} + b^{2}x^{2} = a^{2}b^{2}$. Substitute this value: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{2}}{a^{2}y^{2}} \cdot \frac{a^{2}b^{2}}{a^{2}y}$$Cancel common factors: $$\frac{d^{2}y}{dx^{2}} = -\frac{b^{4}}{a^{2}y^{3}}$$Answer:$$\frac{d^{2}y}{dx^{2}} = \frac{-b^{4}}{a^{2}y^{3}}$$Hence, the correct option is D.